Why generalized force equals to zero?

In summary, the lecturer explains that the generalized coordinate, θ, is equal to q1 and the generalized force, Q1, is equal to 0. The equations for finding partial derivatives and forces in the x and y directions are shown. The conversation then discusses the significance of choosing convenient generalized coordinates and the transformation equations for coordinates. The attempt at a solution shows the process of finding partial derivatives and the resulting generalized force equation, Q(theta), which is equal to 0 due to the constant potential.
  • #1
harmyder
33
1

Homework Statement


Untitled.png

Lecturer says that:
Generalized Coordinate: θ = q1
Generalized Force: Q1 = 0


I don't understand why Q1 equals to zero.

Homework Equations


[itex]Q_i = \sum_j \underline F_j \frac{\delta \underline r_j}{\delta q_i}[/itex]

The Attempt at a Solution


First i need to find partial derivatives:
[itex]\frac{\delta \hat x}{\delta q_1} = \frac{\delta \theta r}{\delta \theta}= r[/itex] (i have lost vector here, but i don't know where to put it in [itex]\theta r[/itex])
[itex]\frac{\delta \hat y}{\delta q_1} = \frac{\delta r}{\delta \theta} = 0[/itex] (same here)

Now forces at x and y directions:
[itex]F_x \frac{\delta \hat x}{\delta q_1} = -F + m\bf g \sin\psi[/itex]
[itex]F_y \frac{\delta \hat y}{\delta q_1} = N - m\bf g \cos\psi[/itex]

Now add them:
[itex]Q_1 = r (-\bf F + m\bf g \sin\psi)[/itex] is it zero?
 
Physics news on Phys.org
  • #2
harmyder said:
Now forces at x and y directions:
Fxδ^xδq1=−F+mgsinψF_x \frac{\delta \hat x}{\delta q_1} = -F + m\bf g \sin\psi
Fyδ^yδq1=N−mgcosψ

what are F and N ?
if they are forces acting on the body and if its frictional and normal reaction forces then check its values ?
 
  • #3
drvrm said:
what is F and N
Friction and reaction forces. You can see it form the image.
 
  • #4
then the magnitude of N should be equal to the component of weight in the direction of normal to the inclined plane and something similar for F .
what is the equation of motion in generalized coordinate , generalized velocities and time-...what is the advantage of such a description?
see
Although there may be many choices for generalized coordinates for a physical system, parameters which are convenient are usually selected for the specification of the configuration of the system and which make the solution of its equations of motion easier. If these parameters are independent of one another, the number of independent generalized coordinates is defined by the number of degrees of freedom of the system.
 
  • #5
My question is why Q1 equals to zero, I'm aware of how to choose generalized coordinates. Thanks.
 
  • #6
Actually generalized forces are transformed from the transformation relations operating between the two sets of coordinates - If a generalized potential can be defined. in that situation a constant potential will mean generalized forces to be zero - in your case the constraining equation is of the type say dx - r. d(theta) =0
In using Lagrangian method the forces of constraint do not appear as the virtual displacements are such that they are consistent with the forces of constraint.
What is the transformation equation of coordinates in your example?
 
  • Like
Likes harmyder
  • #7
drvrm said:
What is the transformation equation of coordinates in your example?

harmyder said:

The Attempt at a Solution


First i need to find partial derivatives:
[itex]\frac{\delta \hat x}{\delta q_1} = \frac{\delta \theta r}{\delta \theta}= r[/itex] (i have lost vector here, but i don't know where to put it in [itex]\theta r[/itex])
[itex]\frac{\delta \hat y}{\delta q_1} = \frac{\delta r}{\delta \theta} = 0[/itex] (same here)

So, x = r*theta and y = r.
 
  • #8
as r is a constant only theta is your generalized coordinate with variation
so delta/delta(theta) = d/dx. dx/dtheta + d/dy. dy/d(theta)

Q(theta) = F(x).r =0
 
  • #9
drvrm said:
...
Sorry, could you please use itex tag, it is hard to understand your equations. Thanks.
 

FAQ: Why generalized force equals to zero?

1. What is generalized force?

Generalized force is a concept in physics that refers to the force acting on a system that is not associated with a specific physical displacement. It is a generalization of the concept of force, which is typically defined as the product of mass and acceleration. Generalized force can include forces such as friction, tension, or torque.

2. Why is it important to understand generalized force?

Understanding generalized force is important because it allows us to analyze the behavior of complex systems, such as those in thermodynamics and quantum mechanics. It also helps us to better understand the relationship between forces and motion in these systems.

3. How is generalized force related to work?

Generalized force is related to work through the concept of virtual work. Virtual work is the work done by a generalized force on a system that undergoes a virtual displacement, which is a hypothetical displacement that does not actually occur. The relationship between generalized force and virtual work is given by the principle of virtual work, which states that the virtual work of all generalized forces acting on a system must equal zero for the system to be in equilibrium.

4. Why does generalized force sometimes equal zero?

Generalized force can equal zero for a variety of reasons. In some cases, the system may be in equilibrium, meaning that all forces acting on it balance each other out. In other cases, the generalized forces may be acting in such a way that their virtual work cancels out, resulting in a net virtual work of zero.

5. How does the concept of generalized force apply to different systems?

The concept of generalized force can be applied to a wide range of systems, including mechanical systems, thermodynamic systems, and quantum systems. In each case, the specific forces and their corresponding virtual work may be different, but the overall principles and relationships remain the same.

Back
Top